Answer
$
\begin{aligned}
\vec{E}_A & =-1.80 \mathrm{~N} / \mathrm{C} \hat{\mathrm{i}}
\end{aligned}
$
Work Step by Step
For point $A$, we have (in SI units)
$
\begin{aligned}
\vec{E}_A & =\left[\frac{q_1}{4 \pi \varepsilon_0 r_1^2}+\frac{q_2}{4 \pi \varepsilon_0 r_2^2}\right](-\hat{\mathrm{i}})=\frac{\left(8.99 \times 10^9\right)\left(1.00 \times 10^{-12} \mathrm{C}\right)}{\left(5.00 \times 10^{-2}\right)^2}(-\hat{\mathrm{i}})+\frac{\left(8.99 \times 10^9\right)\left|-2.00 \times 10^{-12} \mathrm{C}\right|}{\left(2 \times 5.00 \times 10^{-2}\right)^2}(+\hat{\mathrm{i}}) \\
& =(-1.80 \mathrm{~N} / \mathrm{C}) \hat{\mathrm{i}}
\end{aligned}
$
